Q.Find the sum up to the $17^{th}$ term of the series $\dfrac{1^3}{1}+\dfrac{1^3+2^3}{1+3}+\dfrac{1^3+2^3+3^3}{1+3+5}+\cdots$.

Beyond the general AP/GP/AGP sum formulas, three specific finite sums recur so often that they are worth memorising outright:

The square-sum formula can be re-derived from the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ applied telescopically to $k^3-(k-1)^3$, and the cube-sum formula from $k^4-(k-1)^4=4k^3-6k^2+4k-1$ summed telescopically and solved for $\sum k^3$ once $\sum k$ and $\sum k^2$ are already known.

Why these matter beyond the formulas themselves. A great many "compute the sum of first $n$ terms" problems that do not look like an AP or GP reduce to one of these three sums once simplified — for instance, a series whose $k^{th}$ term is $\dfrac{1^3+2^3+\cdots+k^3}{1+3+\cdots+(2k-1)}$ simplifies (using $\sum k^3=\left(\tfrac{k(k+1)}2\right)^2$ in the numerator and $\sum(2j-1)=k^2$ in the denominator) to the clean quadratic $\dfrac{(k+1)^2}4$, whose sum over $k=1,\ldots,N$ is then just a shifted sum-of-squares. Recognising a series in disguise as $\sum k$, $\sum k^2$ or $\sum k^3$ (possibly after algebraic simplification of the general term) is the key skill this section trains.